Hess's Law
Many chemical reactions are difficult to carry out in the lab under normal conditions. The reasons for this are
varied but might include:
- the reaction might be very slow and take a long time to complete.
- there may be unwanted side reactions taking place
This means that we cannot measure the enthalpy changes for these reactions directly. Luckily
we can use Hess's law to
calculate the enthalpy changes for these reactions.
Hess's Law is simply an extension of the law of conservation of energy;
that is that energy cannot be created or destroyed only
change from one form to another. As an example consider the image opposite. Here the change A→B is an exothermic
reaction with an enthalpy change (ΔH1) of -1000kJ.
Now imagine that for some reason the change A→B cannot be
carried out directly. However the changes A→C and C→B are possible with enthalpy changes
of ΔH2= -400kJ and
ΔH3= -600kJ. This route from A→B simply goes through an intermediate C. However the important
fact is that both reactions, the direct route
A→B and the other route through intermediate C both start and end up at the same place.
They both start at A and end up at B.
In essence this is Hess's law. Hess's law
states that:
The enthalpy change for a reaction depends only on the
initial and final states of the reaction and is independent
of the route by which the reaction occurs.
So in our example:
- the direct route A→B has an enthalpy change of ΔH1= -1000kJ.
- the indirect route A→C has an enthalpy change of ΔH2= -400kJ and C→B
has an enthalpy change of ΔH3= -600kJ.
Or we could write ΔH1 = ΔH2 + ΔH3.
That is the sum of the two enthalpy changes, ΔH2 + ΔH3 is the same as the
direct
route ΔH1. Which is just another way of saying that the enthalpy change for a reaction is equal to the sum of the
enthalpy changes for any individual steps, as long as you start and end points
are the same. That's all there is to Hess's law!
Worked examples using Hess' law
Examiners often ask you to calculate the enthalpy change for a reaction using either enthalpies of combustion (ΔcHo)
or enthalpies of formation(ΔfHo). If you are
not sure what these enthalpy changes represent then click the links to do a little research before continuing.
Using enthalpies of combustion (ΔcHo) to calculate an enthalpy change
| substance |
Enthalpy of combustion/kJmol-1 |
| C(s) |
-394 |
| H2(g) |
-286 |
| CH4(g) |
-890 |
Example 1 - Calculate the enthalpy of
formation of methane (CH4(g)) using the enthalpies of combustion given in the table opposite.
All the questions you are likely to be asked to solve are all done the same way. So follow the method below and you can't
go wrong!
Step 1 - You must write the equation for the reaction you are being asked to calculate the
enthalpy change for. In this case we
have to calculate the enthalpy of formation of methane. An equation for
this enthalpy change is given below:
C(s) + 2H2(g) → CH4(g)
To calculate this enthalpy change we have been given enthalpies of combustion data. To be able to work out the enthalpy change you need the enthalpies of combustion
of all the elements and compound present
in the equation above; a quick
check in the table shows that you have all these enthalpies of combustion.
Now all we need to do is draw a simple energy cycle using the data we have.
Some students get confused on how to start these cycles, but if you are using enthalpies of combustion;
which we are here then the cycle
goes "downwards" for the equation above. This is set out below:
Using enthalpies of formation (ΔfHo) to
calculate an enthalpy change
The other type of problem you are likely to meet is where you have to use enthalpies of formation
to calculate an enthalpy change.
| substance |
Enthalpy of formation/kJmol-1 |
| NH3(g) |
-46 |
| NO(g) |
90 |
| H2O(l) |
-286 |
As before the first step is to write out the equation for the reaction you have been asked to find the
enthalpy change for.
Example 2- Calculate the enthalpy change for the following reaction using the
enthalpy of formation(ΔfHo) data in the table.
4NH3(g) + 5O2(g) → 4NOg) + 6H2O(l)
As before in order to calculate the enthalpy of reaction (ΔrxnH) using the enthalpy of formation data provided we need to
know the enthalpy of formation (
(ΔfH) for all the substances in our equation. Looking at the table we have all the
information we need, remember
the enthalpy of formation of an element is by definition 0.
When we drew the energy cycle above using enthalpies of combustion (ΔcHo) we drew the cycle
"downwards" from our equation. This time using enthalpies of formation (ΔfHo) we
will work "upwards" towards our equation.
Key Points
The hardest part of getting started with a Hess's law energy cycles is
knowing where to start - remember that
if you are given enthalpies of combustion data to calculate an unknown
enthalpy change your energy cycle goes "downwards" e.g.
- Using enthalpies of combustion to calculate the enthalpy change for a reaction (ΔHrxn)
- Using Hess’s law we can say that: ΔHrxn = ΔH1 -ΔH2
If you are given enthalpies of formation data to calculate an unknown
enthalpy change your energy cycle
goes "upwards" e.g.
- Using enthalpies of formation to calculate the enthalpy change for a reaction (ΔHrxn)
- Using Hess’s law we can say that:
-
ΔHrxn = -ΔH1 + ΔH2
Or
- ΔHrxn = ΔH2 - ΔH1
Practice questions
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